Buffers, Acid-Base Titration & Ksp Solubility

Buffers, Acid-Base Titration & Ksp Solubility I. Buffers 1. 2. What is a buffer – a solution that is resistant to changes in pH when small amounts of acid or base are added to it How is a buffer solution different from other solutions – buffer solution is composed of a weak acid (HA) and its salt – the conjugate base (A-) or a weak base (B) and its salt conjugate acid (BH+). Acidic buffer: HA (aq) Basic buffer: B (aq) + H2O H+ + ABH+ + OH- 3. Buffer behavior is explained by Le Chatelier’s Principle. 4. Buffer capacity It is a measure of its effectiveness. It is dependent on the amount of weak acid & conjugate base present or amount of weak base and conjugate acid present. A buffer ceases to be a buffer solution when one of its components is removed. Buffer solution calculations A. Determining the pH of a buffer solution 1. 2. Determine the concentrations of each of the buffer components. Solve for the [H+] either using an ICE set-up or the Henderson-Hasselbalch equation to determine the buffer’s pH. pH = pKa + log [base component]/ [weak component] B. Determining the pH of a buffer solution after addition of an acid or a base to the system: This is a two-step problem: stoichiometric and equilibrium problem For example: If you are added 2.00 mLs of 0.10 M HCl to 20.00 mLs of an acetic acid (0.45 M) /sodium acetate (0.55 M) buffer Write an equation for the addition of the acid to the buffer system Stoichiometric Part (moles) H+ + Ac- HAc Before (mols) 0 1.1 x10-2 9.0 x 10-3 addition 2 x 10-4 -2 x 10-4 +2 x 10-4 after 0 1.08 x 10-2 9.2 x 10-3 Note: Total volume is now 22.00 mLs (20.00 mLs + 2.00 mLs). You need to know this information in order to determine the NEW molarities of your buffer components (i.e. weak acid and conjugate or weak base and conjugate acid). Equilibrium Part: (M): Set up an ICE table to determine the pH as given below or at this point in the problem you may use the Henderson-Hasselbalch equation. HAc H3O+ + H2O (l) + Ac- Initial (M) 0.42 0 0.49 Change -x x x Equil 0.42 - x x 0.49 + x Assume x <<<0 .42 Ka = [H3O+] [Ac-] / [HAc] 1.8 x 10-5 = (x) (0.49) / (0.42) 1.54 x 10-5 = x = [H+] pH = 4.8 Check if assumption correct, use 5% rule: 1.54 x 10-5 = x (x / 0.42) x 100 = % (1.54 x 10-5 /0.42) x 100 = 3.7 x10-3 % 3.7 x10-3 % is <<< than 5% assumption is good II. Acid/Base titration Problems A. Terms and Areas on a Titration Graph 1. Equivalence point – point where the moles of base or acid added equals the moles or base of solution being titrated; also known as the stoichiometric point. a. From the equivalence point can determine the molecular mass of acid or base that was titrated (M x V = moles) 2. 1/2 neutralization point – point where the moles of weak acid in solution = its conjugate base and where pH = pKa or the point where the moles of weak base in solution = its conjugate acid and where pOH = pKb 3. buffer area – where the addition of acid or base being added (i.e. Volume is increasing pH is constant) and the pH remains constant. 4. endpoint – color change in the titration; it is the visual equivalence point. 5. titrant – the acid or base in the buret. 6. indicator – chemical substance, usually a weak organic acid that changes colors at certain pH values. The color changes over pH range known as transition interval. Example methyl orange goes from red → orange → yellow in the range of 3.1 to 4.4 B. Strategy for solving titration problems 1. Titrating a strong acid with a strong base. • The simplest type forming water: Equation: H3O+ + OH-  → 2 H2O • Example: If HCl titrated with NaOH a. Before the base is added – the initial pH is the determined using pH = -log [H3O+] b. The equivalence point has a pH of 7: a neutral salt has been formed. • Before the equivalence point is reached a. Determine the moles of H3O+ ion that you must begin with (M X V = moles) b. Determine the moles of OH- ion added (M x v = moles) to that point. c. Determine moles of acid neutralize • The moles of base added x stoichiometric factor = moles of acid neutralized o (initial #moles of H3O+) – (# moles H3O+ neutralized) • concentration of H3O+ ions remaining: o (# moles of H3O+ now neutralize)/ (total volume of solution) * ▪ *(ml of acid to begin with + ml of base added to this point) • Determine pH • At the equivalence point: the pH is 7 – a neutral salt has been produced. • After the equivalence point a. Calculate the volume of OH- in excess i. (Amount of OH- (ml) add) – (amount OH- (ml) at the equivalence point) ii. excess OH- (L) x M of NaOH = moles of OHiii. determine total volume of solution (see *) b. Calculate pOH c. Calculate pH 2. Titrating a weak acid with a strong base: the equivalence point is GREATER that pH of 7 • The initial pH of the weak acid is determined using the Ka of the weak acid a. Similar to a weak acid equilibrium problem calculation. b. Before the equivalence point, the solution ‘behaves’ like a buffer. c. NOTE: A buffer solution IS NOT FORMED. The solution exhibits a buffering like behavior because of the presence of the weak acid & conjugate base species. d. The pH at the half-neutralization point can be determined using the H/H equation or an ICE table. • At the ½ neutralization point the weak acid [HA] = [A-] its conjugate base. • Other points on the titration curve prior to reaching the equivalence point can be determined as you would determine the pH of a buffer as if base is being added to the solution. • At the equivalence point: a. The weak acid [HA] has been completely neutralized by the added base. • Moles of weak acid [HA] in solution now equal the moles of base added. • The pH at this point is totally due to the presence of the conjugate base [A-] that is in solution. • The pH at the equivalence point is determined using the Kb not Ka of the weak acid. • The moles of conjugate base (A−) = the moles of HA (original). • The concentration (M) of the conjugate base in solution is (moles of A -)/ total volume. • Total volume is the starting volume + the milliliters of base added to reach the equivalence point. • The equilibrium reaction for this point is A- + H2O (l)  → HA + OH• The pH of the solution is determined using an ICE table. • To determine the pH of the solution beyond the equivalence point” a. The OH− is in excess. • [OH−] = moles of hydroxide ions/total volume (liters) • -log [OH−] = pOH • pH + pOH = 14 3. Titrating a weak base with a strong acid: • The initial pH of the weak base is determined using the Kb of the weak base. a. Similar to a weak base equilibrium problem calculation. b. Before the equivalence point, the solution ‘behaves’ like a buffer. c. NOTE: A buffer solution IS NOT FORMED. The solution exhibits a buffering like behavior because of the presence of the weak acid & conjugate base species. d. The pH at the half-neutralization point can be determined using the H/H equation or an ICE table. • At the ½ neutralization point the weak acid [B] = [BH+] its conjugate acid. • Other points on the titration curve and prior to reaching the equivalence point can be determined as you would determine the pH of a buffer. • At the equivalence point: a. The pH of the equivalence point for a weak base: the pH is less than 7. Reason: salt formed is an acidic salt. b. The weak base [B] has been completely neutralized by the added acid. c. Moles of weak base [B] in solution now equal the moles of acid added. • The pH at this point is totally due to the presence of the conjugate acid [BH +] that is in solution. d. The pH at the equivalence point is determined using the Ka not Kb of the weak base. • The moles of conjugate acid (BH+) = the moles of original weak base (B). • The concentration (M) of the conjugate acid in solution is (moles of BH +)/ total volume. • Total volume is the starting volume + the milliliters of acid added to reach the equivalence point. e. The equilibrium reaction for this point is BH+ + H2O (l)  → B + H3O+ f. The pH of the solution is determined using an ICE table. • To determine the pH of the solution beyond the equivalence point” a. The H3O+ (acid) is in excess. b. [H3O+] = moles of hydronium ions/total volume (liters) c. -log [H3O+] = pH III. Ksp and Common Ion Solubility • Ksp looking at the solubility of a substance under specific conditions • The Ksp expression says if excess solid is added to the system it will not influence the POSITION of the solubility’s equilibrium A. Comparing Ksp • Ksp of salts can only be compared directly if the dissolution of the salts has the same ratio – same number of ions B. Ksp Problems Types • Two types: Calculating Ksp given the concentration of the species Calculating solubility (s) for all chemical species when given Ksp value. C. Predicting Ksp and the Ion product Q Ksp: • tells us the amount of solid that can be dissolved in each volume of water at a specific temperature • tells the amount of solid needed for a precipitate • is not about equilibrium position but about equilibria • Ion product Q: similar to the equilibrium Q value that we used to predict whether or not a system is at equilibrium; the ion product Q helps predict whether or not something will precipitate out of solution or not. • If Q = Ksp this tells you the maximum amount of solid that can be dissolved in a given amount of solvent without precipitation – a saturated solution • If Q is greater than Ksp then have precipitation • If Q is less than Ksp will not have precipitation – unsaturated solution.